Showing the Polar form of the Cauchy-Riemann Equations imply the Cartesian Form Often times, when the Cauchy-Riemann equations $$u_x = v_y; \ u_y = -v_x$$ are introduced, it's an exercise to show that one can arrive at the "polar form" of those equations. That is, the cartesian Cauchy Riemann equations imply the equations $$r \cdot \frac{\partial{u}}{\partial{r}} = \frac{\partial{v}}{\partial{\theta}}; \ \frac{\partial{u}}{\partial{\theta}} = -r \frac{\partial{v}}{\partial{r}}$$
However, if I am to call these second equations the Polar form of the Cauchy-Riemann equations, I'd like there to be an equivalence of the two sets of equations. That is, I would like the polar equations to imply the Cartesian ones as well. However, a proof of this converse is often completely missing from many texts, and the converse proof doesn't seem to be as immediately obvious (at least to me).
Naively, one could try and replicate the same style of proof often given for the forward direction. Namely, we could use chain rule to deduce $$u_x = u_r \cdot r_x + u_{\theta} \cdot \theta_x$$ and do the same for $v_y$, but explictly computing $r_x, \theta_x, r_y, \theta_y$ is admittedly quite a bit trickier and involves dividing by various trig functions or taking inverse trig functions of things and then computing derivatives (at least with the methods I've tried). In either case, we now have to be much more careful making sure every expression is well-defined for values of $\theta$.  So far, I've not been able to attain the Cartesian Cauchy-Riemann equations, so perhaps a little help in the right direction would be greatly appreciated.
 A: Let's try it this way:
$$\begin{aligned} 
u_r &= u_x x_r + u_y y_r = u_x \cos \theta +u_y \sin \theta\\
v_r &= v_x x_r + v_y y_r = v_x \cos \theta +v_y \sin \theta
\end{aligned}$$
$$\begin{aligned} 
u_\theta &= u_x x_\theta + u_y y_\theta =   u_y r\cos \theta -u_x r \sin \theta\\
v_\theta &= v_x x_\theta + v_y y_\theta =   v_y r\cos \theta -v_x r \sin \theta
\end{aligned}
$$
$$\begin{aligned} 
r \, u_r &= v_\theta \\
u_x \cos \theta +u_y \sin \theta &= v_y \cos \theta -v_x  \sin \theta\\
(u_x - v_y) \cos \theta +(v_x+u_y) \sin \theta &= 0
\end{aligned}
$$
$$\begin{aligned} 
-r \, v_r &= u_\theta \\
-v_x \cos \theta -v_y \sin \theta &= u_y \cos \theta -u_x  \sin \theta\\
(u_x-v_y) \sin \theta -(v_x + u_y) \cos \theta &= 0
\end{aligned}
$$
Combining the last equation in the third and fourth "sets" of equations:
$$ \pmatrix{(u_x-v_y)\cos \theta +(v_x+u_y) \sin \theta \\ -(v_x+u_y)\cos \theta+ (u_x-v_y) \sin \theta}=\pmatrix{0\\0}$$
or
$$ \pmatrix{\cos \theta & \sin \theta\\ -\sin\theta & \cos \theta}\pmatrix{u_x-v_y\\v_x+u_y}=\pmatrix{0\\0}.$$
Since the matrix on the left is a rotation matrix, it must be that the vector $\pmatrix{u_x-v_y\\v_x+u_y}$ is identically zero.
$$\bbox[5px, border: 1pt solid blue]{\begin{aligned}u_x&=\phantom{-}v_y,\\u_y&=-v_x.\end{aligned}}$$
Notice, in the above, that we've assumed $r\ne 0,$  otherwise, there is not much information in the polar form of the C-R equations.  There (at the origin), both $u_\theta$ and $v_\theta$ are zero.  Actually, at $z=0$, the parameter $\theta$ is anyway not well defined.
